Consider a planet with the same mass and diameter as Earth, except that it is

1) perfectly spherical.

2) a hollow shell 1 meter thick, so the planet is almost like a massive ping-pong ball.

Naturally it would be made of some super-dense material such that it had the same mass as earth without most of its volume.

If you stood on the surface of this planet, the gravity should be equivalent to Earth's, but if you were to drill a hole in the surface and climb through, would gravity:

a) pull you toward the planet's center of mass, which is in the empty center?

b) draw you toward the interior surface, and thus outward from the planet's center, such that you could walk along the inside of the planet?

c) somewhere in between, with little or no gravitational tendency?

I am posting this in Physics but I am not sure if World building is more appropriate.


The gravitational force you would feel inside the hollow planet is zero. Let's prove it. Let's call the interior of the planet $P$ (it is an open ball in $\mathbb{R}^3$.

1) Gravitational force is conservative, which means that $F = - \nabla \Phi$, where $F$ is the gravitational force, and $\Phi$ is a smooth function (the potential).

2) By Gauss' Theorem for gravity, $\text{div} F = -4 \phi G \rho$, where $\rho$ is the mass density at a point. Since, in $P$, $\rho = 0$, $\text{div} F = 0$ in $P$.

3) By spherical symmetry, any spherical shell $S \subset P$, having the same center as $P$ is equipotential (otherwise you'd have a force field which is not spherically symmetric).

4) Therefore, $\Phi$ satisfies the equation $\text{div}\nabla \Phi = \Delta \Phi = 0$, and $\Phi = \text{const}$ on the sphere $S$. This means that $\Phi = \text{const}$ inside $S$, which implies $F = 0$.

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    $\begingroup$ For extra bonus points, consider the fact that Earth is a rotating sphere. If this hollow earth was rotating, you would be able to walk on the inside of the sphere, although the force appearing to hold you there would be quite weak. $\endgroup$ – Floris Jan 29 '16 at 23:47
  • $\begingroup$ I wasn't considering that, but I suppose that's a factor. $\endgroup$ – PlasmaStarfish Jan 29 '16 at 23:52

C. Newton's Shell theorem states that if you're inside a hollow spherical shell that has matter even distributed in the shell, then the net gravitational pull you'd feel would be 0.

This is because the attraction to the small near part of the shell is equal to the attraction to the large far part of the shell.


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